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I am looking to prove that $f:E_1 \rightarrow E_2$ is continuous iff $f^{-1}(V)$ is open whenever $V$ is open.

Assuming continuity, I was able to show that the inverse image of any open set in the codomain is open in the domain. It's the other direction I am struggling with. Here is the set up:

Let $V$ be an open set in the metric space $E_2$ such that $f^{-1}(V)=U$ is open in the metric space $E_1$. Also let $p_0\in E_1$ and $\epsilon >0$ be given. We need to specify some value $\delta >0$ that forces $f(p) \in B_{E_2}[f(p_0),\epsilon]$ whenever $p \in B_{E_1}[p_0, \delta]$.

The proof I am looking at continues by suggesting that since $B_{E_2}[f(p_0),\epsilon]$ is open, the inverse image is open by assumption, and then since $p_0 \in f^{-1}(B_{E_2}[f(p_0),\epsilon])$, there is a radius value $r>0$ that forces $B_{E_1}[p_0, r]$ to be a subset of $f^{-1}(B_{E_2}[f(p_0),\epsilon])$ (by the definition of openness). So for any $p \in B_{E_1}[p_0, r]$, we have $f(p) \in B_{E_2}[f(p_0,\epsilon]$ as required and are done.

I am confused as to why we can initially say that $f(p_0)$ is in $V$. What if, for example, the generic $p_0$ is not in $U=f^{-1}(V)$? Any help on this point would be appreciated.

EDS
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    I think $V \subseteq E_2$ and $U \subseteq E_1$ are generic open sets. So your first statement shouldn't really be "Let $V$ be an open set in the metric space E_2 such that....", but it should be "Suppose for every open $V \subseteq E_2$ the preimage $f^{-1}(V)=U$ is open in $E_1$. Then proceed – Jihoon Kang Apr 20 '19 at 17:34

2 Answers2

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Take a point $p_0$ in $E_1$. We would like to show that $f$ is continuous at $p_0$.

Your approach really is basically correct. $f(p_0)$ is a point in $E_2$. For any $\epsilon > 0$, write $V = B_{E_2}[f(p_0),\epsilon]$. This is an open set, so $f^{-1}(V)$ is an open set in $E_1$. Note that $U = f^{-1}(V)$ contains $p_0$, since $f(p_0)$ is in $V$ by definition.

Because $U$ is open, there is some $\delta > 0$ such that the ball $B_{E_1}[p_0,\delta]$ is contained in $U$. But since $U = f^{-1}(V)$ this precisely says that whenever $d_{E_1}(p_0,p) < \delta$, $d_{E_2}(f(p_0),f(p)) < \epsilon$.

Rylee Lyman
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    Ah, so I needed to define the open set based on the point. Open balls are open sets, and by the assumption, the inverse of ANY open set in the codomain is open in the domain. Thanks! – EDS Apr 20 '19 at 19:26
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This holds in the general setting of topological spaces (not necessarily metrizable).

Assume that for every open subset $V$ of $E_2,$ $f^{-1}(V)$ is an open subset of $E_1.$

Let $p_0\in E_1.$ We want to prove that $f$ is continuous at $p_0,$ i.e. (by definition) that

for every open subset $V$ in $E_2$ containing $f(p_0),$ there exists an open subset $U$ in $E_1$ containing $p_0$ and such that $f(U)\subset V.$

Well... simply take $U=f^{-1}(V).$

As for the converse (which you said you did succeed to prove), see also Prove that the inverse image of an open set is open.

Anne Bauval
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